Question

If $\alpha=\sin \left(\frac{\sin ^{-1} \frac{1}{\sqrt{3}}}{3}\right)$ $\beta=\cos \left(\cos ^{-1}\left(\frac{1}{\sqrt{5}}\right)-\sin ^{-1}\left(\frac{2}{\sqrt{5}}\right)\right)$ then $\frac{\beta^{2}}{\left(3 \alpha-4 \alpha^{3}\right)^{2}}$ is equal to

Answer 1

Let $\sin ^{-1} \frac{1}{\sqrt{3}}=\theta$
$\Rightarrow \sin \theta=\frac{1}{\sqrt{3}}$
$\therefore \alpha=\sin \frac{\theta}{3}$
Hence, $\sin \left(3 \cdot \frac{\theta}{3}\right)=3 \sin \frac{\theta}{3}-4 \sin ^{3} \frac{\theta}{3}=3 \alpha-4 \alpha^{3}$
$\therefore 3 \alpha-4 \alpha^{3}=\sin \theta=\frac{1}{\sqrt{3}}$
Also, $\beta=\cos \left(\cos ^{-1} \frac{1}{\sqrt{5}}-\sin ^{-1} \frac{2}{\sqrt{5}}\right)$
$=\cos \left(\tan ^{-1} 2-\tan ^{-1} 2\right)=\cos \left(0^{\circ}\right)=1$
$\therefore \frac{\beta^{2}}{\left(3 \alpha-4 \alpha^{3}\right)^{2}}=\frac{1}{\left(\frac{1}{\sqrt{3}}\right)^{2}}=3 .$